Step-by-step Solution:

We are given the condition: \[ \prod_{i=1}^{n} \tan (\alpha_i) = 1, \quad \forall \alpha_i \in \left[0, \frac{\pi}{2} \right] \] We need to find the maximum value of: \[ \prod_{i=1}^{n} \sin (\alpha_i). \] Step 1: Express \(\tan(\alpha_i)\) in Terms of \(\sin(\alpha_i)\) and \(\cos(\alpha_i)\)
Using the identity: \[ \tan(\alpha_i) = \frac{\sin(\alpha_i)}{\cos(\alpha_i)}, \] we can rewrite the given condition as: \[ \prod_{i=1}^{n} \frac{\sin (\alpha_i)}{\cos (\alpha_i)} = 1. \] Rearranging: \[ \prod_{i=1}^{n} \sin (\alpha_i) = \prod_{i=1}^{n} \cos (\alpha_i). \] Step 2: Find Maximum Value
We define: \[ f(\alpha_1, \alpha_2, ..., \alpha_n) = \prod_{i=1}^{n} \sin (\alpha_i), \] subject to the constraint: \[ \prod_{i=1}^{n} \sin (\alpha_i) = \prod_{i=1}^{n} \cos (\alpha_i). \] Taking the natural logarithm on both sides: \[ \sum_{i=1}^{n} \ln \sin (\alpha_i) = \sum_{i=1}^{n} \ln \cos (\alpha_i). \] Using AM-GM inequality, the maximum occurs when all \( \alpha_i \) are equal, i.e., \[ \alpha_1 = \alpha_2 = \dots = \alpha_n = \alpha. \] Thus, we solve: \[ \tan^n \alpha = 1 \Rightarrow \tan \alpha = 1 \Rightarrow \alpha = \frac{\pi}{4}. \] Substituting this into our target expression: \[ \prod_{i=1}^{n} \sin (\alpha_i) = \sin^n \frac{\pi}{4} = \left(\frac{\sqrt{2}}{2}\right)^n = 2^{-\frac{n}{2}}. \] Final Answer: \[ {2^{-\frac{n}{2}}} \]